find lim x→0 sin²(2x) / 3x². as your answer, please enter the value of the limit in decimal form with three…

Explanation:

Step1: Use the double - angle formula

We know that $\sin(2x)=2\sin x\cos x$, so $\sin^{2}(2x) = 4\sin^{2}x\cos^{2}x$. Also, we use the well - known limit $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$. Let $u = 2x$. As $x\rightarrow0$, $u\rightarrow0$. We rewrite the given limit $\lim_{x\rightarrow0}\frac{\sin^{2}(2x)}{3x^{2}}$ as $\lim_{x\rightarrow0}\frac{4\sin^{2}(2x)}{12x^{2}}$.

Step2: Apply the limit formula

$\lim_{x\rightarrow0}\frac{\sin^{2}(2x)}{3x^{2}}=\frac{4}{3}\lim_{x\rightarrow0}(\frac{\sin(2x)}{2x})^{2}$ Since $\lim_{u\rightarrow0}\frac{\sin u}{u} = 1$, when $u = 2x$ and $x\rightarrow0$, $\lim_{x\rightarrow0}\frac{\sin(2x)}{2x}=1$.

Step3: Calculate the limit value

$\frac{4}{3}\times1^{2}=\frac{4}{3}\approx1.333$

Answer:

$1.333$